40048 + Interview Questions in Physics in Class 11 Page 272 InterviewSolution

13551.	In which of the following cases the centre of mass of a rod is certainly not at its centre (A) The density continuously increases from left to right (B) The density continuously decrease from left to right(C ) The density decreases from left to right upto centre and then increases(D) The density increases from left to right upto centre and then decreases
Answer» A & B A & C B & D C & D Answer :A

Discussion

13552.	A plano convex lens fits exactly in to a plano concave lens. Their plane surfaces are parallel to each other. IF the lenses are made of different materials of refractive indices mu_1 and mu_2 and R is the radius of curvature of the curved surfaces of the lenses when the focal length of the combination is
Answer» `R/(mu_1-mu_2)` `(2R)/(mu_1-mu_2)` `R/(2(mu_1-mu_2))` `R/(mu_1+mu_2)` ANSWER :A

Discussion

13553.	A: A thread of length L is streatched by a force of 5N, its length increases by a and stretched by force of 6N its length increases by b, now if it is stretched by force of 11 N its length increases by (a + b -L). R: Increasing the length of elastic thread is proportional to its initial length.
Answer» Both are true and the reason is the correct explanation of the ASSERTION. Both are true but the reason is not correct explanation of the assertion. Assertion is true, but the reason is false. Both assertion and reason are false. Solution :If initial length is L and increase in length l due to force F then in `l = (FL)/(AY), (L)/(AY)=C` `l = FC` `therefore ` For first case ` a = L +l` `therefore a = L + FC = L + 5 C ""…(1)` `therefore b = L + F.C = L + 6C""…(2)` For second case, `therefore b -a = L + 6C -L-5C` `therefore b -a =C` If increase in length is x due to force 11N then, `x = L +11C""[F=11x]...(3)` From EQUATION (1) and (2), `C = (a -L)/(5) and C = (b-L)/(6)` `therefore (a-L)/(5) = (b-L)/(6)` `therefore 6a - 6L = 5-5L` `therefore 6a 05b =L""...(4)` Putting value of equation (4) in equation (3), `x = 6a - 5b + 11( b-a) [because C = b -a]` `=6a - 5b +11b -11a` `x = 6b -5a` `therefore` Assertion is false and reason in true.

Discussion

13554.	In Melde's experiment , when a string is stretched by a piece of glass it vibrates with7 loops . When the glass piece is completely immersed in water the string vibrates in9 loops. What is the specific gravity of glass ?
Answer» <P> SOLUTION :If `F_(1) and F_(2)` are the TENSIONS in the strings in the two cases , then `F_(1) p_(1)^(2) = F_(2) P_(2)^(2)` `(F_(1))/(F_(2)) = (P_(2)^(2))/(P_(1)^(2)) = ( 9^(2))/( 7^(2)) = ( 81)/( 49)` If `F_(b)` is the buoyant force on the GLASS piece , then ` F_(b) = F_(1) - F_(2)` Specific gravity of glass ` = ( "Weight in AIR" )/( "Loss in weight in water" )` ` = (F_(1))/( F_(b)) = (F_(1))/( F_(1) - F_(2)) = (1)/( 1 - (F_(2))/(F_(1)))` `(1)/( 1- ( 49)/( 81)) = 2.53`

Discussion

13555.	A weight W rests on a rough horizontal plane. If the angle of friction is theta, the least force that can move the body along the plane will be
Answer» W cos `theta` W TAN `theta` W cot `theta` W SIN `theta` Answer :D

Discussion

13556.	linear expansion is change in length of an object with temperature. Write the equation for cofficient of linear expansion.
Answer» SOLUTION :`alpha_l=(TRIANGLEL)/(L TRIANGLE T )`

Discussion

13557.	Inside a uniform sphere of mass M(M is mass of complete sphere) and radius R, a cavity of radius R/3is madeas shown. Find the escape velocity of a particle projected from point A.
Answer» SOLUTION :`SQRT((88GM)/(45R))`

Discussion

13558.	Side of the regular hexagon is 2 meter. Magnetic field at point O due to the network shown in the figure sqrt3B_(0)xx10^(-7) T. Find value of B_(0)
Answer» ANSWER :2

Discussion

13559.	A simple pendulum is carrying a bob of mass 'M'. If is pulled through an angle 45^(@) from its mean position. The horizontal force required to keep it under equilibrium at this new position is
Answer» MG Mg/2 `Mg//SQRT(2)` `sqrt(2)Mg` ANSWER :A

Discussion

13560.	Assertion:The time period of a pendulum on a satellite orbiting the Earth is infinity.Reasons:Time period of a pendulum is inversely proportional to square root of acceleration due to gravity.Which of the following statements is correct?
Answer» Both assertion and reason are true and reason explain assertion correctly. Both assertion and reason are true and reason does not explain assertion correctly. Assertion is true but reason is false. Assertion is false and reason ALSO false. Solution :Assertion:In a satellite, acceleration due to gravity g = 0 `T=2pisqrt((L)/(0))=infty` Reason:The time PERIOD of a SIMPLE PENDULUM is`T=2pisqrt((l)/(g))`

Discussion

13561.	Is it possible that the average velocity for some interval may be zero although the average velocity for a shorter interval included in the first interval is not zero?
Answer» Solution :Yes. If a particle moves ALONG a straight line with constant 1 st in one direction say in + x direction for some interval of time `Delta t_(1)` then it everses its direction and moves for another time interval `Delta t_(2)` and reaches the same POINT then for total interval `Delta t_(1)+Delta t_(2)` the average velocity is ZERO `(:. (Deltax)/(Deltat)= "O"/(Deltat)=0)`displacement is zero , but for 1st or 2 nd time INTERVALS the average velocity is not zero. Is `(Deltax)/(Deltat)`, the average velocity of the particle during a time interval `vecDeltat` is equal to the slope of the staight line joining initial and final points on the position - time graph.

Discussion

13562.	A metal sheet with a circular hole is heated. The hole will
Answer» contract EXPAND remain unaffected contract of expand DEPENDING on the VALUE of the LINER EXPANSION coefficient. Solution :expand

Discussion

13563.	Four moles of a perfect gas heated to increase its temperature by 2^@C absorbs heat of 40 cal atconstant volume. If the same gas is heated at constant pressure the amount of heat supplied is (R = 2 cal/mol K)
Answer» 28 CAL 56 cal 84 cal 94 cal Answer :B

Discussion

13564.	A Fahrenheat thermometer reads 113^(@) F while a faulty celsius thermometer reads 44^(0)C. The correction to be applied to the celsius thermometer is
Answer» `-1^(0) C ` `+ 1^(0) C ` `+ 2^(@) C ` `-2^(0)` C Answer :B

Discussion

13565.	Two rods of different materials are matained at the same temperature different at their ends. Their lenghts are l_(1),l_(2) cross sectionalareas are A_(1),A_(2)and coeffcients of thermal conductivity are K_(1),K_(2) . If the heat at the steady state is same in both the rods, what condition should be satisfied ?
Answer» Solution :Rate of FLOW of heat means Q/t., Given`(Q)/(t) = (K_(1)A_(1)(d THETA))/(l_(1)) = (K_(2)A_(2)(d theta))/(l_(2)) rArr (l_(1))/(K_(2)A_(2))`, i.e.,their thermal resistanceare the sameIf `l_(1)- l_(2)`then `K_(1)A_(1) = K_(2)A_(2) ,`If `A_(1) = A_(2) ` then`K_(1)//K_(2) = l_(1)//l_(2)`

Discussion

13566.	Find the vector product of vectors vecA = 4hati + 2 hatJ - hatk ,vecB = hati+ 3 hatJ + 4 hatk
Answer» `3 hati -htj-5hatk` `11 hati -17htj+10hatk` `4 hati +6htj-4hatk` `5 hati + 5htj+3hatk` Solution :`11 hat(I )-17hat(j )+ 10 hat(k )` `= (hat(I ) (8+3) - hat(j )(16 + 1) + hat(k )(12-2)` `11 hat(I )-17 hat(j )+ 10 hat(k )`

Discussion

13567.	If n drops of liquid, each with surface energy E, join to form a single drop, then
Answer» some energy will be RELEASED in the PROCESS some energy will be ABSORBED in the process the energy released or absorbed will be `E( n-n^(2//3))` the energy released or absorbed will be `nE (2^(2//3)–1)` Answer :A::C

Discussion

13568.	The motion of a particle is given by X = A sin omegat + Bcos omegat the motion of the particle is
Answer» not SIMPLE harmonic simple harmonic with AMPLITUDE A+B simple harmonic with amplitude `(A+B)//2` simple harmonic with amplitude `sqrt(A^(2)+B^(2))` ANSWER :D

Discussion

13569.	(A) : The method of dimensions cannot be used to derive the formula N=N_(0) c^(- lambda I). (R) : Exponential functions have no dimensions.
Answer» Both (A) and (R) are true and (R) is the CORRECT explanation of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is false Both (A) and (R) are false Answer :A

Discussion

13570.	Given relative error in the measurement of length is 0.02, what is the percentage error?
Answer» ANSWER :`2%`

Discussion

13571.	Statement I: All the rain drops hit the surface of the earth with the same constant velocity. Statement II: An object falling through a viscous medium eventually attains a terminal velocity.
Answer» Statement I is true, statement II is true , statement II is a CORRECT EXPLANATION for statement I. Statement I is true, statement II is true , statement II is not a correct explanation for statement I. Statement I is true, statement II is false. Statement I is false, statement II is true. Answer :A

Discussion

13572.	(A) : A shell moving in a parabolic path explodes in mid air. The centre of mass of the fragments will follow the same parabolic path.(R ) : Explosion is due to internal forces, which cannot alter the state of motion of centre of mass.
Answer» Both (A) and (R ) are true and (R ) is the CORRECT EXPLANATION of (A) Both (A) and (R ) are true and (R ) is not the correct explanation of (A) (A) is true but (R ) is FALSE Both (A) and (R ) are false ANSWER :A

Discussion

13573.	Explain work done by torque.
Answer» Solution :Suppose a rigid body rotating about a fixed axis, which is taken as the Z-axis. This axis is perpendicular to plane X.Y.. Let force `vec(F_(1))` acting on a particle of the body at point `P_(1)` and it rotates on a circle of radius `r_(1)` with centre C on the axis `CP_(1)=r_(1)`. In TIME `Deltat`, the point MOVES to the position `P._(1)` from `P_(1)`. The displacement of the particle is `DeltaS_(1)=r_(1)Deltatheta` and it is in the direction tangential at `P_(1)therefore Deltatheta=angleP_(1)OP._(1)` is the angular displacement of the particle. The WORK done by the force `vec(F_(1))` on the particle is `dW_(1)=vec(F_(1)).dvec(S_(1))` `=F_(1)DeltaS_(1)cosphi_(1)` `=F_(1)r_(1)Deltathetacos(90^(@)-alpha_(1))` where `DeltaS_(1)=r_(1)Delta thetaand phi_(1)+alpha_(1)=90^(@)` where `phi_(1)` is the angle between `vec(F_(1))` and the tangent at `P_(1) and alpha_(1)` is the angle between `vec(F_(1))` and the radius vector `vec(OP)=vec(r_(1)), phi` The torque due to `vec(F_(1))` about the origin `vectau=vec(OP_(1))xxvec(F_(1))""....(1)` but ACCORDING to figure (b) `vec(OP)_(1)=vec(OC)+vec(CP)_(1)` Since `vec(OP)` is along the axis, `therefore TAU=OC(F_(1))sin0^(@)=0`, hence `vec(OP)_(1)=vec(CP)_(1)` From eqn. (1), `vec(tau_(1))=vec(CP)_(1)xxvec(F_(1))` `therefore vec(tau_(1))=vec(r_(1))xxvec(F_(1))[because vec(CP_(1))=vec(r_(1))]` `therefore tau_(1)=r_(1)F_(1)sinalpha_(1)` and work `DeltaW_(1)=tau_(1)Deltatheta` If there are more than one forces acting on the body the work done by all of them can be added to give the total work done on the body `DeltaW=(tau_(1)+tau_(2)+tau_(3)+....tau_(n))Deltatheta` `therefore DeltaW=tauDeltatheta""....(2)` where `tau_(1)+tau_(2)+tau_(3)+....tau_(n)=tau` total torque If `Deltathetararr0` total work done, `dW=taud theta` is similar to eqn. `dW=Fds` in linear motion Diving (2) on both in eqn. (2) `(DeltaW)/(Deltat)=tau.(Deltatheta)/(Deltat)` If `Deltatrarr0=(DeltaW)/(Deltat)=(dW)/(dt)and(Deltatheta)/(Deltat)=(d theta)/(dt)` `therefore` Instantaneous power, `P=tau.(d theta)/(dt)` `therefore P=tauomega` This is the instantaneous power. It is similar to equation `P=Fv` in linear motion.

Discussion

13574.	Due to the applicaton of a force on a body of mass 100 kg that is initially at rest, The body moves with an accleration of 20ms^(-2) in the direction of the force . Find the magnitude of the force .
Answer» ANSWER :2000N

Discussion

13575.	If g on the surface of the earth is 9.8m//s^2, find its value at a height of 6400 km. (Radius of the earth = 6400km)
Answer» Solution :LET g at a HEIGHT H be `g_h` `g_h=g/(1+h/R)^2` h=6400 KM, R=6400 km, `g=9.8 m//s^2` `g_h=9.8/(1+6400/6400)^2=9.8/(1+1)^2 =9.8/4=2.45 "m//s"^2`

Discussion

13576.	Select the correct alternative: When a conservative force does positive work on a body, the potentail energy of the body increases/ decreases/remains unaltered.
Answer» SOLUTION :DECREASES

Discussion

13577.	A stone is dropped freely, while another thrown vertically downward with an initial velocity of 2 "ms"^(-1)from the same point, simultaneously. The time required by them to have a distance of separation 22 m between them is
Answer» 11s 5.5s 44s 22s Answer :A

Discussion

13578.	Find the units of mass and length if the unit of force is Mega Newton, unit of power is watt and unit of times is 1 microsecond.
Answer» Solution :`M= 10^(6) KG, L= 10^(-12)m`

Discussion

13579.	Refer to the previous problem. Suppose, the goldsmith argues that he has not mixed copper or any other material with gold, rather some cavities might have been left inside the ornament. Calcuate the weighs given in that problem.
Answer» Solution :Given `m_(Au)=36gm` `-Again ((m_(Au))/(rho_(Au))+V_C)rhoPw+g=2xxg` ` [where `V_c=`VOLUME of CAVITY]` `RARR (36/19.3+V_c)xx1=2` `rarr V_c=2-36/19.3` `=(36-6)/19.3` =0.112cm^3`

Discussion

13580.	A 10kW drilling machine is used for 5 minutes to bore a hole in an aluminium block of mass 10 xx10^(3) kg. If 40% of the work done is utilised to raise the temperature of the block,find the rise in temperature of the aluminium block ?
Answer» Solution :Work done by the drilling MACHINE in 5 minutes =W `W = "POWER" xx "time" = 10 xx10^(3) xx5xx60 = 3 xx10^(6) J` The energy utilised to RAISE the temperature of the block=40 % of W =`3 xx10^(6) xx 40/100 = 12 xx10^(5) J`. Heat gained by aluminium block = mass `xx`specific heat `xx` increase in temperature. `12 xx10^(5) =(10xx10^(3))xx0.9xx Delta t ` ` :.Delta t = (12XX10^(5)/0.9xx10^(4)) = 133.3^(@)C`

Discussion

13581.	In the pulley system shown in figure. P and Q are fixed pulleys while A, B and C are movable pulleys each of mass 1kg. The strings are vertical and inextensible. Find the tension in the string and acceleration of frictionless pulleys A, B and C
Answer» Solution :A single string whose ENDS are tied to CENTRES of A and B, passes over all the pulleys, so tension at each point of string is same equal to T. Weight of each pulley A, B and C is mg = 1 g Newton. Let `y_A ,y_B and y_C` be the distances of centres of pulleys A, B and C from fixed pulleys at any time t. Following the string starting from end A and REACHING upto end B, we have `(y_B - y_A) + y_B + 2y_A + y_C + y_C- y_B = L`= constant i.e.,` y_A + y_B + 2y_C = L = ` contant DIfferentitating twice with respect to t , we get ` (d^(2)y_A)/(dt^(2)) + (d^(2)y_B)/(dt^(2)) + 2(d^(2)y_C)/(dt^(2)) =0` i.e., `a_A + a_B + 2a_C =0` ...... (1) where ` a_A , a_Band a_C` are acceleration of pulley A,B,C respectively . Now equations of motion of pulleys A,B and C are `mg + T-2T = ma_A implies mg-T =ma_A`............ (2) `mg + T-2T = ma_B implies mg -T = ma_B`......... (3) and `mg-2T = ma_C` From (2) and (3) it is obvious that and ` a_A = a_B = (g-(T)/(M)) `............ (5) and from (4) , `a_C = g-(2T)/(m)`............. (6) susbtituting `a_A , a_B and a_C` in (1) , we get `(g-(T)/(m)) + (g-(T)/(m)) + 2(g-(2T)/(m)) =0` ` 4G- (6T)/(m) =0 implies T = (2)/(3) mg = (2)/(3) xx 1 xx 9.8 = 6.5 N` `:. a_A =a_B = (g-(T)/(m)) = 9.8 -(6.5)/(1) = 3.3 m//s^(2)` From (1) , `ac=-a_A = -3.3 m//s^(2)`

Discussion

13582.	A ball hits a floor normally with velocity u and return with same speed. If the ball is in contact with the floor for 0.1 sec, find the average impulsive force acting on the ball. Take the mass of the ball m = 200 gm and u=2m//s.
Answer» Solution :FORCE on the BALL : `"Average impulsive force, "` `VECF=(DELTAVECP)/(Deltat)rArr""F=(2m u)/(Deltat)=(2(0.2)2)/(0.1)=8N` Hence, force on of the ball is 8 N upwards. The force on the FLOOR by the ball is 8 N downwards.

Discussion

13583.	A spring of force constant 'K' is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of
Answer» `(2K)/3` `(3K)/2` 3k 6k Answer :B

Discussion

13584.	The gravitational field in a region is given by vecE = (4hati +3hatj)N/Kg. The gravitational potential at the points (3m, 0) and (0, 4m). If the potential at the origin is taken to be zero.
Answer» SOLUTION :`VECV=-(vecE.vecr)` At (3m,0) `BARV=-(4hati+3hatj).(3HATI)`=-12J/kg At (0,4 m) `vecV=-(4hati+3hatj).(4hatj)`=-12J/kg

Discussion

13585.	A body of mass m is placed on a platform of mass M ( m lt M), moving with a velocity v. If the coefficient of friction between the platform and the mass is mu then for how long will the body continue to slide on the platform and what distance will it cover during that time?
Answer» SOLUTION :When the platform moves the body on it tends to slide backwards. So the FRICTIONAL force on thebody acts in the direction of motion of the platform. It is given by , f = `mu`mg. `:.` Its acceleration a = `(f)/(m) = mu g` When the velocity of the body becomes v it no longer slides over the platform. We know v = u+at `"""or", t = (v+u)/(a)` Here u=0, a = `mu g :. t = (v)/(mu g)` Again `v^(2) = u^(2) +2 as """or", s = (v^(2)-u^(2))/(2a) = (v^(2))/(2mu g)` `:.` The body will slide for a TIME of `(v)/ (mu g)`, and in this interval of time it will cover a distance of`(v^(2))/(2mug).`

Discussion

13586.	In which of the following cases the centre of mass of a rod is certainly not at its centre (A) The density continuously increases from left to right (B) The density continously decreases from left to right (C ) The density decreases from left to right upto centre and then increases (D) The density increases from left to right upto centre and then decreases
Answer» A & B A & C B & D C & D Answer :A

Discussion

13587.	A position dependent force f=(7-2x+3x^(2))N acts on a small body of mass 2 kg and displaces it from x=0 " to " x=5m. Work done is
Answer» (a) 35J (B) 70J (c) 135J (d) 270J Solution :`W=underset(x_(1))overset(x_(2))intFdx=underset(0)overset(5)INT(7-2x3x^(2))dx=[7x-x(2)+x^(3)]_(0)^(5)=135J`

Discussion

13588.	One lighi year is.........
Answer» `3.153 xx 10^(7)` m `1.496 xx 10^(7)` m `9.46 xx 10^(12)` KM `3.26 xx 10^(15)` m Solution :`9.46 xx 10^(12)` km

Discussion

13589.	The difference between volume and pressure coefficients of an ideal gas is ……………… .
Answer» `(1)/(273)` 273 `(2)/(273)` zero Answer :d

Discussion

13590.	A solid at temperature T_(1) is kept in an evacuated chamber at temperature T_(2) > T_(1). The rate of increase of temperature of the body is proportional to
Answer» `T_(2) - T_(1) ` `T_(2)^(2) - T_(1)^(2)` `T_(2)^(3) - T_(1)^(3)` `T_(2)^(4) - T_(1)^(4)` ANSWER :D

Discussion

13591.	An elevator whose floor-to-ceiling destance is 2.50 m starts ascending with a constant acceleration of 1.25 ms^(-2) On second after the start, a bolt begins falling from the elevator. Calculate: a. The free fall time of the bolt b. The displacement and reference frame of ground.
Answer» Solution :a. ` t=sqrt((2h)/(g+a))=sqrt((2xx2.5)/(10+1.25))=(2)/(3) s ` B. Velocity of LIFT after `1` s. `v=0 +1.25 xx 1=1.25 ms^(-1) =5//4 ms^(-1)` Ths will be the initial velocity of bolt. Destance moved up by lift in `2//3 s`, ` x=(5)/(4)xx(20/(3)+(1)/(2)1.25((2)/(3))^(2)=(10)/(9)m` Displacement of bolt `=2.5-x=2.5 -(10)/(9) =(25)/(18) m` Maximum height attained by bolt above the point of dropping , `(v^(2))/(2g)=((5//4)^(2))/(2xx 10) =(5)/(64) m` So distance travelled=`2xx (5)/(64)+ (25)/(18)=(445)/(288)m`.

Discussion

13592.	A metal block of mass 10.2 kg is dragged along a rough horizontal surface at a constant speed of 8.6 m/s. If the coefficient of friction between the surface and the block be 0.46 find (i) the power generated and (ii) if 40% of the heat is absorbed by the block, find the rate of rise in temperature. Specific heat capacity of the metal is 126"J/kg"-^(@)C.
Answer» ANSWER :(i) 395.4 W (II) 0.31^(@)C//s`

Discussion

13593.	A point source emits sound equally in all directions in a non-absorbing medium. Two points P and Q are at distance of 2 m and 3 m respectively from the source. The ratio of the intensities of the waves at P and Q is ......
Answer» <P>`3:2` `2:3` `9:4` `4:9` SOLUTION :We have,`I = (P)/(A) = (P)/( A pi R ^(2)) IMPLIES I prop (1)/(r ^(2))` `therefore (I _(1))/( I _(2)) = (r _(2) ^(2))/( r _(1) ^(2)) = ((3) ^(2))/( (2) ^(2)) = 9/4` `therefore I _(1) :I _(2) =9:4`

Discussion

13594.	A plece of ice floats in a beaker containing water with a certain fraction inside water. If the density of ice is D and that of water is rho find the value of n. What will happen to the level of water when the ice completely melts ?
Answer» Solution :Volume of ice` XX` DENSITY of ice = Volume of imeresed PORTION of ice` xx` density of water ` "" VD = n V rho ,n= D//rho ` Volume of ice INSIDE water = `n V = (DV)/( rho )` mass of ice = VD = mass of water formed Volume of water formed `= ("Mass")/("Density") = (VD)/(rho)` This means when the whole ice melts, the volume of water formed is EQUAL to the volume displaced by ice. So the level of water will not rise.

Discussion

13595.	A stationary source emits sound of frequncy f_(0) = 492 Hz. The sound is reflected by a large car approaching the source with a speed of 2 ms^(-1). The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz ? (Given that the speed of sound in air is 330 ms^(-1) and the car reflects the sound at the frequency it has received.) {:(0,1,2,3,4,5,6,7,8,9),(,,,,,,,,,):}
Answer» Solution :For reflected sound we can assume it to be a case of source and observer approaching towards each other. If U is speed of car and `UPSILON` is speed of sound then FREQUENCY received after reflection can be WRITTEN as follows : `f = f_(0) ((upsilon + u)/(v - u))` Beat frequency can be written as follows : `Delta f = f_(0) ((upsilon + u)/(upsilon - u)) - f_(0)` `rArr""Delta f = f_(0) (2u)/(upsilon - u)` Substituting values we get the following : `rArr""Delta f = 492 xx ((2 xx 2)/(330 - 2)) = 6`

Discussion

13596.	Mark the correct options (A) If the far point goes ahead, the power of the divergent lens should be reduced (B) IF the near point goes ahead, the power of convergent lens should be reduced ( C) IF the far point is 1m away from the eye, divergent lens should be used (D) IF the near point is 1m away from the eye, divergent lens should be used.
Answer» (A) (C ) and (D) (A) and (D) (A) (D) and ( C) (A) and ( C) ANSWER :D

Discussion

13597.	If angular velocity of a rIgId body at 10 cm distance from rotational axis is 10 rad/s, then the linear velocity of particle is …... .
Answer» ANSWER :`50cm//s` `v=vecomegaxxvecr` `=omegarsintheta` `=omegarsin90^@` `=OMEGAR` `:. v=10xx5` `:.v=50cm//s`

Discussion

13598.	The number of significant figures in 6.023xx10^(23)"mole"^(-1) is
Answer» 4 3 2 23 Answer :A

Discussion

13599.	The efficiency of a cannot's engine is 20%. When the temperature of the source is increased by 25^(@)C, then its efficiency is found to increase to 25%. Calculate the temperature of source and sink.
Answer» Solution :Temperature of source `= T_(1)` Temperature of sink `= T_(2)` EFFICIENCY `= eta = 20%=0.2` `eta = 1 - (T_(2))/(T_(1)), (T_(2))/(T_(1)) = 1 - 0.2 = 0.8 i.e., T_(2) = 0.8 T _(1)` When the temperature of the source is INCREASED by `25^(@)C,` the new efficiency is 25%. `T_(2)` is same. `0.25 = 1 - (T_(1))/(T_(1)+_25) , i.e, (T_(2))/(T_(1) + 25) = 1 - 0.25 = 0.75` `T_(2) = 0.75 (T_(1) + 25) = 0.74 T_(1) + 0.75 XX 25` But ` T_(2) = 0.8 T_(1)` `therefore 0.8 T _(1) = 0.75 T_(1) + 18. 75 ` `0.5 T_(1) = 18.75 ,T_(1) = (18.75)/(0.05) = 375K` `T_(2) = 0.8 T_(1) = 0.8 xx 375= 300 K`

Discussion

13600.	The bob of a pendulum is made of a hollow brass sphere. What happens to the time period of the pendulum. If the bob is filled with water completely? Why?
Answer» Solution :If the hollow brass SPHERE in COMPLETELY filled with water then time period of simple PENDULUM does not change. This is because time period of a pendulum is INDEPENDENT of MASS of the bob.

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